On 12/07/2016 11:36 am, smitra wrote:
On 12-07-2016 02:00, Bruce Kellett wrote:
On 12/07/2016 2:56 am, smitra wrote:
On 11-07-2016 13:49, Bruce Kellett wrote:
On 11/07/2016 9:31 pm, Bruno Marchal wrote:
HOLIDAY EXERCISE:
A guy undergoes the Washington Moscow duplication, starting again
from Helsinki.
Then in Moscow, but not in Washington, he (the one in Moscow of
course) undergoes a similar Sidney-Beijing duplication.
I write P(H->M) the probability in H to get M.
In Helsinki, he tries to evaluate his chance to get Sidney.
With one reasoning, he (the H-guy) thinks that P(H-M) = 1/2, and
that P(M-S) = 1/2, and so conclude (multiplication of independent
probability) that P(H-S) = 1/2 * 1/2 = 1/4.
But with another reasoning, he thinks that the duplications give
globally a triplication, leading eventually to a copy in W, a copy
in S and a copy in B, and so, directly conclude P(H-S) = 1/3.
So, is it 1/4 or 1/3 ?
Neither. The probability that the guy starting from Helsinki gets to
Sydney is unity. This is the problem with probabilities in the MWI --
how do you interpret probabilities when all possible outcomes occur
with probability one?
Bruce
In duplication experiments the prior probability to exist at all in
any of the possible states increases after the duplication, while in
unitary QM this is conserved (except if one or more of the possible
outcomes is death). The correct way to analyze Bruno style
duplication arguments is to start with assigning some measure m to
the observer before any duplication is carried out, in this case the
observer at H.
What has a measure got to do with it?
There exists a probability to exist at all, in case of duplication
experiments this is non-trivial, the probability to find yourself
alive increases after duplication.
The probability that you find that you exist is one. This probability is
not increased by duplication. If you do not exist, the probability that
you exist is zero. If you are looking to see if you exist, then the
probability you exist is exactly one -- it cannot be anything else.
Then H gives rises to two copies in states W and M (we can call them
copies, but they are actually different observers as they have
different memories stored in their brains, so they are different
algorithms). The measures will be m for each of these observers.
Then W is not going to be copied, while M gives rise to S and B, so
we end up with 3 observers each with measure m. The probability is
thus 1/3.
If they are different observers (different algorithms) then the only
possible answer for the prior probability is P(H->S) = 0. The original
is destroyed and does not go anywhere.
Yes, but then we're a bit too pedantic, what it means is that you have
different observers which have the same history up to the duplication,
and if they were to go to sleep, wake up and at that moment not
remember where they are, then you have again identical observers for a
few seconds who then will diverge again when they access their memories.
But in principle, you could say that each instant always gives rise to
a new observer, as I'm typing this sentence I'm constantly changing,
so strictly speaking this email was written by many different people.
The closest continuer theory (with new identities in the case of ties)
removes this confusion.
In an analogue MWI setting, the outcome is different, at each
duplication the measure for a particular outcome is halved. W thus
has a measure of m/2, while S and B each have a measure of m/4, the
probability is thus 1/4.
In MWI, P(H->S) = 1. The only way you can get P(H->S) = 1/4 is in a
collapse model: H has to definitely go to M in order for S to become a
possibility. In non-collapse models, H goes to both M and W, so
P(H->M) = 1. Subsequent duplication in the non-collapse model leads to
copies in S and B, with both probabilities equal to one.
To get any probability other than unity or zero, you require either a
collapse or a dualist model. (The dualist model is the implicit
assumption that there is some hidden label whereby we can distinguish
the observers at W, S and B -- observers have to carry a unique
identity with them.) In a more reasonable interpretation of personal
identity in the duplication cases, each duplication creates two new
persons, so the probabilities become P(H->W) = P(H->M) = P(H->S) =
P(H->B) = 0.
MWI yields the same probabilities as other interpretations of QM.
The concept of probability is problematic in MWI. That is why people are
spending so much effort to derive the Born rule from within MWI (to
avoid imposing it from outside). Probabilities only make sense if you
have unique outcomes -- if all possible outcomes are realized, then the
probability for any particular outcome is necessarily one.
Bruce
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at https://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.
